Question 445: Electrostatics & Capacitance

Question

A parallel plate capacitor having plate area ' $A$ ' and separation ' $d$ ' is charged to a potential difference ' $V$ '. The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between the plates is ( $\varepsilon_0=$ permittivity of free space)

Accepted Answer

Step 1: Work Done Formula
The work done to separate the plates is the change in energy of the capacitor:
$$ W = U_{	ext{final}} - U_{	ext{initial}} $$
Step 2: Initial Energy
When the plates are at the original distance, the energy stored is:
$$ U_{	ext{initial}} = \frac{1}{2} C V_0^2 $$

Here, $C$ is the initial capacitance, and $V_0$ is the voltage across the plates.
Step 3: What Happens When the Plates Are Separated
After disconnecting the battery, the charge $Q$ on the plates stays the same, but the distance is increased to $4d$.
The new capacitance is:

$$ C' = \frac{\varepsilon_0 A}{4d} = \frac{C}{4} $$
The voltage changes because $Q$ is the same:

$$ V' = \frac{Q}{C'} = \frac{Q}{C/4} = 4V_0 $$
Step 4: Final Energy
The new stored energy is:

$$ U_{	ext{final}} = \frac{1}{2} C' {V'}^2 = \frac{1}{2} \left(\frac{C}{4}ight) (4V_0)^2 = \frac{1}{2} 	imes \frac{C}{4} 	imes 16V_0^2 = 2CV_0^2 $$
Step 5: Calculate the Work Done
Plugging values into the work formula:

$$ W = U_{	ext{final}} - U_{	ext{initial}} = 2CV_0^2 - \frac{1}{2}CV_0^2 = \frac{3}{2}CV_0^2 $$
Step 6: Substitute Capacitance
The capacitance formula is $C = \frac{\varepsilon_0 A}{d}$. So,

$$ W = \frac{3}{2} \cdot \frac{\varepsilon_0 A}{d} V_0^2 = \frac{3 \varepsilon_0 A V_0^2}{2 d} $$

Answer

$\frac{\varepsilon_0 A V^2}{3 d}$

Answer

$\frac{\varepsilon_0 A V^2}{4 d}$

Answer

$\frac{2 \varepsilon_0 A V^2}{d}$

Answer

$\frac{3 \varepsilon_0 A V^2}{2 d}$

Electrostatics & Capacitance

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